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                <h2 id="「力扣」第-658-题：找到-K-个最接近的元素"><a href="#「力扣」第-658-题：找到-K-个最接近的元素" class="headerlink" title="「力扣」第 658 题：找到 K 个最接近的元素"></a>「力扣」第 658 题：找到 K 个最接近的元素</h2><p>题解地址：<a href="https://leetcode-cn.com/problems/find-k-closest-elements/solution/pai-chu-fa-shuang-zhi-zhen-er-fen-fa-python-dai-ma/" target="_blank" rel="noopener">排除法（双指针） + 二分法（Python 代码、Java 代码）</a>。</p>
<p>说明：文本首发在力扣的题解版块，更新也会在第 1 时间在上面的网站中更新，这篇文章只是上面的文章的一个快照，您可以点击上面的链接看到其他网友对本文的评论。</p>
<p>传送门：<a href="https://leetcode-cn.com/problems/find-k-closest-elements/" target="_blank" rel="noopener">658. 找到 K 个最接近的元素</a>。</p>
<blockquote>
<p>给定一个排序好的数组，两个整数 k 和 x，从数组中找到最靠近 x（两数之差最小）的 k 个数。返回的结果必须要是按升序排好的。如果有两个数与 x 的差值一样，优先选择数值较小的那个数。</p>
<p>示例 1:</p>
<p>输入: [1,2,3,4,5], k=4, x=3<br>输出: [1,2,3,4]</p>
<p>示例 2:</p>
<p>输入: [1,2,3,4,5], k=4, x=-1<br>输出: [1,2,3,4]</p>
<p>说明:</p>
<p>k 的值为正数，且总是小于给定排序数组的长度。<br>数组不为空，且长度不超过 104<br>数组里的每个元素与 x 的绝对值不超过 104</p>
<p>更新(2017/9/19):<br>这个参数 arr 已经被改变为一个整数数组（而不是整数列表）。 请重新加载代码定义以获取最新更改。</p>
</blockquote>
<h2 id="排除法（双指针）-二分法（Python-代码、Java-代码）"><a href="#排除法（双指针）-二分法（Python-代码、Java-代码）" class="headerlink" title="排除法（双指针） + 二分法（Python 代码、Java 代码）"></a>排除法（双指针） + 二分法（Python 代码、Java 代码）</h2><p>做这一类题目的思路往往来自于对具体例子的研究，多举几个例子，在草稿纸上写写画画，也有助于我们对边界问题的讨论。</p>
<p>以下介绍的两种方法，排除法比较容易想到，而<strong>二分法基于排除法的思想</strong>，希望读者能够认真体会，代码虽然简单，但是要做一些分类讨论才能解释得清楚。</p>
<h3 id="方法一：排除法（双指针）"><a href="#方法一：排除法（双指针）" class="headerlink" title="方法一：排除法（双指针）"></a>方法一：排除法（双指针）</h3><p>以 <code>arr = [1, 2, 3, 4, 5, 6, 7]</code> , <code>x = 5</code>, <code>k = 3</code> 为例。 </p>
<p><strong>思路分析</strong>：</p>
<p>1、一个一个删，因为是有序数组，且返回的是连续升序子数组，<strong>所以每一次删除的元素一定是位于边界</strong>；</p>
<p>2、一共 $7$ 个元素，要保留 $3$ 个元素，因此要删除 $4$ 个元素；</p>
<p>3、因为要删除的元素都位于边界，于是可以使用<strong>双指针</strong>对撞的方式确定保留区间，即“最优区间”。</p>
<p>（温馨提示：下面的幻灯片中，有几页上有较多的文字，可能需要您停留一下，可以点击右下角的后退 “|◀” 或者前进 “▶|” 按钮控制幻灯片的播放。）</p>
<p><img src="https://pic.leetcode-cn.com/7cac52f799ba654e9483f47cd17533706a380dbc15fbb6dbc93a83bce11c45db-658-1.png" alt="658-1.png">),<img src="https://pic.leetcode-cn.com/ab9c789d3c7fc18cdb1dd7d383c3fa984033f4420412ffe91d747cdfe3623994-658-2.png" alt="658-2.png">),<img src="https://pic.leetcode-cn.com/0ae58bb7d22e120a900c8eb7a5735ef5e777373885d17f8f90fe0a172579716a-658-3.png" alt="658-3.png">),<img src="https://pic.leetcode-cn.com/3b003257d123fe2e944f0f06f94d88d1b35cc0f071748dc5ca2f77362c69621e-658-4.png" alt="658-4.png">),<img src="https://pic.leetcode-cn.com/9b971313fa81e97893a9321769478375dac74fa92d8cdc75563386949f18731d-658-5.png" alt="658-5.png"></p>
<p><strong>参考代码</strong>：</p>
<p>Java 代码：</p>
<pre class="line-numbers language-java"><code class="language-java"><span class="token keyword">import</span> java<span class="token punctuation">.</span>util<span class="token punctuation">.</span>ArrayList<span class="token punctuation">;</span>
<span class="token keyword">import</span> java<span class="token punctuation">.</span>util<span class="token punctuation">.</span>List<span class="token punctuation">;</span>

<span class="token keyword">public</span> <span class="token keyword">class</span> <span class="token class-name">Solution</span> <span class="token punctuation">{</span>

    <span class="token keyword">public</span> List<span class="token operator">&lt;</span>Integer<span class="token operator">></span> <span class="token function">findClosestElements</span><span class="token punctuation">(</span><span class="token keyword">int</span><span class="token punctuation">[</span><span class="token punctuation">]</span> arr<span class="token punctuation">,</span> <span class="token keyword">int</span> k<span class="token punctuation">,</span> <span class="token keyword">int</span> x<span class="token punctuation">)</span> <span class="token punctuation">{</span>
        <span class="token keyword">int</span> size <span class="token operator">=</span> arr<span class="token punctuation">.</span>length<span class="token punctuation">;</span>

        <span class="token keyword">int</span> left <span class="token operator">=</span> <span class="token number">0</span><span class="token punctuation">;</span>
        <span class="token keyword">int</span> right <span class="token operator">=</span> size <span class="token operator">-</span> <span class="token number">1</span><span class="token punctuation">;</span>

        <span class="token keyword">int</span> removeNums <span class="token operator">=</span> size <span class="token operator">-</span> k<span class="token punctuation">;</span>
        <span class="token keyword">while</span> <span class="token punctuation">(</span>removeNums <span class="token operator">></span> <span class="token number">0</span><span class="token punctuation">)</span> <span class="token punctuation">{</span>
            <span class="token keyword">if</span> <span class="token punctuation">(</span>x <span class="token operator">-</span> arr<span class="token punctuation">[</span>left<span class="token punctuation">]</span> <span class="token operator">&lt;=</span> arr<span class="token punctuation">[</span>right<span class="token punctuation">]</span> <span class="token operator">-</span> x<span class="token punctuation">)</span> <span class="token punctuation">{</span>
                right<span class="token operator">--</span><span class="token punctuation">;</span>
            <span class="token punctuation">}</span> <span class="token keyword">else</span> <span class="token punctuation">{</span>
                left<span class="token operator">++</span><span class="token punctuation">;</span>
            <span class="token punctuation">}</span>
            removeNums<span class="token operator">--</span><span class="token punctuation">;</span>
        <span class="token punctuation">}</span>

        List<span class="token operator">&lt;</span>Integer<span class="token operator">></span> res <span class="token operator">=</span> <span class="token keyword">new</span> <span class="token class-name">ArrayList</span><span class="token operator">&lt;</span><span class="token operator">></span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">;</span>
        <span class="token keyword">for</span> <span class="token punctuation">(</span><span class="token keyword">int</span> i <span class="token operator">=</span> left<span class="token punctuation">;</span> i <span class="token operator">&lt;</span> left <span class="token operator">+</span> k<span class="token punctuation">;</span> i<span class="token operator">++</span><span class="token punctuation">)</span> <span class="token punctuation">{</span>
            res<span class="token punctuation">.</span><span class="token function">add</span><span class="token punctuation">(</span>arr<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token punctuation">)</span><span class="token punctuation">;</span>
        <span class="token punctuation">}</span>
        <span class="token keyword">return</span> res<span class="token punctuation">;</span>
    <span class="token punctuation">}</span>

    <span class="token keyword">public</span> <span class="token keyword">static</span> <span class="token keyword">void</span> <span class="token function">main</span><span class="token punctuation">(</span>String<span class="token punctuation">[</span><span class="token punctuation">]</span> args<span class="token punctuation">)</span> <span class="token punctuation">{</span>
        <span class="token keyword">int</span><span class="token punctuation">[</span><span class="token punctuation">]</span> arr <span class="token operator">=</span> <span class="token punctuation">{</span><span class="token number">0</span><span class="token punctuation">,</span> <span class="token number">0</span><span class="token punctuation">,</span> <span class="token number">1</span><span class="token punctuation">,</span> <span class="token number">2</span><span class="token punctuation">,</span> <span class="token number">3</span><span class="token punctuation">,</span> <span class="token number">3</span><span class="token punctuation">,</span> <span class="token number">4</span><span class="token punctuation">,</span> <span class="token number">7</span><span class="token punctuation">,</span> <span class="token number">7</span><span class="token punctuation">,</span> <span class="token number">8</span><span class="token punctuation">}</span><span class="token punctuation">;</span>
        <span class="token keyword">int</span> k <span class="token operator">=</span> <span class="token number">3</span><span class="token punctuation">;</span>
        <span class="token keyword">int</span> x <span class="token operator">=</span> <span class="token number">5</span><span class="token punctuation">;</span>
        Solution solution <span class="token operator">=</span> <span class="token keyword">new</span> <span class="token class-name">Solution</span><span class="token punctuation">(</span><span class="token punctuation">)</span><span class="token punctuation">;</span>
        List<span class="token operator">&lt;</span>Integer<span class="token operator">></span> res <span class="token operator">=</span> solution<span class="token punctuation">.</span><span class="token function">findClosestElements</span><span class="token punctuation">(</span>arr<span class="token punctuation">,</span> k<span class="token punctuation">,</span> x<span class="token punctuation">)</span><span class="token punctuation">;</span>
        System<span class="token punctuation">.</span>out<span class="token punctuation">.</span><span class="token function">println</span><span class="token punctuation">(</span>res<span class="token punctuation">)</span><span class="token punctuation">;</span>
    <span class="token punctuation">}</span>
<span class="token punctuation">}</span><span aria-hidden="true" class="line-numbers-rows"><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span></span></code></pre>
<p>Python 代码：</p>
<pre class="line-numbers language-python"><code class="language-python"><span class="token keyword">from</span> typing <span class="token keyword">import</span> List


<span class="token keyword">class</span> <span class="token class-name">Solution</span><span class="token punctuation">:</span>
    <span class="token keyword">def</span> <span class="token function">findClosestElements</span><span class="token punctuation">(</span>self<span class="token punctuation">,</span> arr<span class="token punctuation">:</span> List<span class="token punctuation">[</span>int<span class="token punctuation">]</span><span class="token punctuation">,</span> k<span class="token punctuation">:</span> int<span class="token punctuation">,</span> x<span class="token punctuation">:</span> int<span class="token punctuation">)</span> <span class="token operator">-</span><span class="token operator">></span> List<span class="token punctuation">[</span>int<span class="token punctuation">]</span><span class="token punctuation">:</span>
        <span class="token comment" spellcheck="true"># 排除法（双指针）</span>
        size <span class="token operator">=</span> len<span class="token punctuation">(</span>arr<span class="token punctuation">)</span>
        left <span class="token operator">=</span> <span class="token number">0</span>
        right <span class="token operator">=</span> size <span class="token operator">-</span> <span class="token number">1</span>

        <span class="token comment" spellcheck="true"># 我们要排除掉 size - k 这么多元素</span>
        remove_nums <span class="token operator">=</span> size <span class="token operator">-</span> k
        <span class="token keyword">while</span> remove_nums<span class="token punctuation">:</span>
            <span class="token comment" spellcheck="true"># 调试语句</span>
            <span class="token comment" spellcheck="true"># print(left, right, k)</span>
            <span class="token comment" spellcheck="true"># 注意：这里等于号的含义，题目中说，距离相等的时候取小的</span>
            <span class="token comment" spellcheck="true"># 所以，相等的时候，尽量缩小右边界</span>
            <span class="token keyword">if</span> x <span class="token operator">-</span> arr<span class="token punctuation">[</span>left<span class="token punctuation">]</span> <span class="token operator">&lt;=</span> arr<span class="token punctuation">[</span>right<span class="token punctuation">]</span> <span class="token operator">-</span> x<span class="token punctuation">:</span>
                right <span class="token operator">-=</span> <span class="token number">1</span>
            <span class="token keyword">else</span><span class="token punctuation">:</span>
                left <span class="token operator">+=</span> <span class="token number">1</span>
            remove_nums <span class="token operator">-=</span> <span class="token number">1</span>
        <span class="token keyword">return</span> arr<span class="token punctuation">[</span>left<span class="token punctuation">:</span>left <span class="token operator">+</span> k<span class="token punctuation">]</span><span aria-hidden="true" class="line-numbers-rows"><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span></span></code></pre>
<p><strong>复杂度分析</strong>：</p>
<ul>
<li>时间复杂度：$O(N)$，这里 $N$ 是数组的长度。</li>
<li>空间复杂度：$O(1)$，只使用了常数个额外的辅助空间。</li>
</ul>
<hr>
<p>题目中说有序数组，又易知：  </p>
<p>1、题目要求返回的是区间，并且是连续区间；  </p>
<p>2、区间长度是固定的，并且 <code>k</code> 的值为正数，且总是小于给定排序数组的长度，即 <code>k</code> 的值“不违规”；  </p>
<p>因此，只要我们找到了左边界的索引，从左边界开始数 <code>k</code> 个数，返回就好了。我们把这件事情定义为“寻找最优区间”，“寻找最优区间”等价于“寻找最优区间的左边界”。因此本题使用二分查找法在有序数组中<strong>定位含有 <code>k</code> 个元素的连续子区间的左边界</strong>，即使用二分法找“最优区间的左边界”。</p>
<h3 id="方法二：二分查找最优区间的左边界"><a href="#方法二：二分查找最优区间的左边界" class="headerlink" title="方法二：二分查找最优区间的左边界"></a>方法二：二分查找最优区间的左边界</h3><p>由排除法，我们知道：</p>
<p><strong>“排除法”的结论</strong>：（这个结论对于这道问题来说非常重要，可以说是解题的关键）</p>
<blockquote>
<p>如果 <code>x</code> 的值就在长度为 size 区间内（不一定相等），要得到 size - 1 个符合题意的最接近的元素，此时看左右边界：</p>
<p>1、如果左边界距离 <code>x</code> 较近，删除右边界；<br>2、如果右边界距离 <code>x</code> 较近，删除左边界；<br>3、如果左、右边界距离 <code>x</code> 的长度相等，删除右边界。</p>
</blockquote>
<p><strong>讨论“最优区间的左边界”的取值范围</strong>：</p>
<p>首先我们讨论左区间的取值范围，使用具体的例子，就很很清楚地找到规律：  </p>
<p>1、假设一共有 5 个数，<code>[0,1,2,3,4]</code>，找 3 个数，左边界最多到 2；  </p>
<p>2、假设一共有 8 个数，<code>[0,1,2,3,4,5,6,7]</code>，找 5 个数，左边界最多到 3。</p>
<p>因此，“最优区间的左边界”的索引的搜索区间为 <code>[0, size - k]</code>，注意，这个区间的左右都是闭区间，都能取到。</p>
<p>定位左区间的索引，有一点技巧性，但并不难理解。由排除法的结论，我们先从 <code>[0, size - k]</code> 这个区间的任意一个位置（用二分法就是当前候选区间的中位数）开始，<strong>定位一个长度为 <code>(k + 1)</code> 的区间</strong>，根据这个区间是否包含 <code>x</code> 开展讨论。</p>
<p>1、如果区间包含 <code>x</code>，我们尝试删除 1 个元素，好让区间发生移动，便于定位“最优区间的左边界”的索引；<br>2、如果区间不包含 <code>x</code>，就更简单了，我们尝试把区间进行移动，以试图包含 <code>x</code>，但也有可能区间移动不了（极端情况下）。</p>
<p>以下的讨论，对于记号 <code>left</code>、<code>right</code> 和 <code>mid</code> 说明如下：</p>
<p>1、<code>left</code>、<code>right</code> 是候选区间的左右边界的索引，根据上面的分析，初始时，<code>left = 0</code>，<code>right = size - k</code>；<br>2、而 <code>mid</code> 是候选区间的中位数的索引，它的取值可能是</p>
<pre><code>mid = left + (right - left) // 2</code></pre><p>也可能是</p>
<pre><code>mid = left + (right - left + 1) // 2</code></pre><p>之所以我们选择 <code>mid = left + (right - left) // 2</code> ，请参考我在「力扣」第 35 题：搜索插入位置的题解<a href="https://leetcode-cn.com/problems/search-insert-position/solution/te-bie-hao-yong-de-er-fen-cha-fa-fa-mo-ban-python-/" target="_blank" rel="noopener">《特别好用的二分查找法模板（Python 代码、Java 代码）》</a>中的叙述。</p>
<blockquote>
<p>后面的文字可能会非常绕，在这里建议读者通读，前后来回看，不太清楚的地方先跳过，且不一定全看我的叙述，看明白一小段，在草稿纸上写写画画一点，卡壳了再看我的叙述，这样就不会太晕。</p>
</blockquote>
<p>我们先从最简单的情况开始讨论：</p>
<p>1、如果区间不包含 <code>x</code>：</p>
<p>（1） 区间的右端点在 <code>x</code> 的左边，即 <code>x</code> 比 <code>arr</code> 中最大的元素还要大，<strong>因为要去掉 1 个元素，显然去掉左端点</strong>，因此“最优区间的左边界”的索引至少是 <code>mid + 1</code>，即 <code>left = mid + 1</code>，<strong>因为区间不可能再往左边走了</strong>，如图；</p>
<p><img src="https://pic.leetcode-cn.com/002e341fa376ece19580704839a5a8bad78b50c6c93a148a928b840ea8cd0272-image.png" alt="image.png"></p>
<p>说明：极端情况是此时中位数位于索引 <code>size - k</code>，区间不能右移。</p>
<p>（2）区间的左端点在 <code>x</code> 的左边，即 <code>x</code> 比 <code>arr</code> 中最小的元素还要小，当前的区间左端点的索引至多是 <code>mid</code>，此时 <code>right = mid</code>，<strong>因为区间不可能再往右偏了</strong>，如图；</p>
<p><img src="https://pic.leetcode-cn.com/4fe43ad19083c07fb72771892f36cb1d5b0dba01533e522f075ef0a153a1267a-image.png" alt="image.png"></p>
<p>说明：极端情况是此时 <code>mid</code> 位于索引 <code>0</code>，区间不能左移。</p>
<p>2、如果区间包含 <code>x</code>，我们尝试删掉一个元素，以便让区间发生移动，缩小搜索范围：</p>
<p>易知，我们要比较长度为 <code>k + 1</code> 的区间的左右端点的数值与 <code>x</code> 的距离。此时这个区间的左边界的索引是 <code>mid</code>，右边界的索引是 <code>mid + k</code>。根据“排除法”的结论，分类讨论如下：</p>
<p>（1）如果右边界距离 <code>x</code> 较近，左边界收缩，可以肯定的是“最优区间的左边界”的索引 <code>left</code> 至少是 <code>mid + 1</code>，即 <code>left = mid + 1</code>，如图；</p>
<p><img src="https://pic.leetcode-cn.com/0a8fe24c6abcad7ae2d774506b4b0abccd5eb95b953be9c367f49c633fed9343-image.png" alt="image.png"></p>
<p>说明：“右边界距离 <code>x</code> 较近”同样适用于 1、（1）情况，因此它们二者可以合并；</p>
<p>（2）如果左边界距离 <code>x</code> 较近，右边界收缩，此时区间不移动，注意：此时有可能收缩以后的区间就是待求的区间，也有可能整个区间向左移动，这件事情叫做，<code>right = mid</code> 不能排除 <code>mid</code>，如图；</p>
<p><img src="https://pic.leetcode-cn.com/6dfea16f88fd03e10c95c2e2d216711ed1489a58a173ac4cd1d2e1a9de583de0-image.png" alt="image.png"></p>
<p>说明1：这一点比较难想，但实际上也可以不想，根据 2、（1）的结论，左区间收缩的反面即是右区间不收缩，因此，这一分支的逻辑一定是 <code>right = mid</code>。</p>
<blockquote>
<p>“实际上也可以不想”的具体原因，同样参考我在「力扣」第 35 题：搜索插入位置的题解<a href="https://leetcode-cn.com/problems/search-insert-position/solution/te-bie-hao-yong-de-er-fen-cha-fa-fa-mo-ban-python-/" target="_blank" rel="noopener">《特别好用的二分查找法模板（Python 代码、Java 代码）》</a>中的叙述，我专门把如何写好二分法，使用二分法模板好用的地方、使用它的技巧和注意事项整理在这篇题解中，希望能对大家有所帮助。</p>
</blockquote>
<p>说明2：“左边界距离 <code>x</code> 较近”同样适用于 1、（2）情况，因此它们二者可以合并。</p>
<p>（3）如果左、右边界距离 <code>x</code> 的长度相等，删除右边界，结论同 2、（2），也有 <code>right = mid</code>，可以合并到 2、（2）。</p>
<p>以上看晕的朋友们，建议你在草稿纸上写写画画，思路就非常清晰了，并且写出的代码也很简洁。这个代码也不是我原创的，在网上搜了一下，刚开始的时候，一直不能理解下面这段代码的意思。</p>
<pre><code>if x - arr[mid] &gt; arr[mid + k] - x:
    left = mid + 1
else:
    right = mid</code></pre><p>写个草稿就清楚多了，原来是并不困难，只是稍显复杂。</p>
<p><img src="https://pic.leetcode-cn.com/1505e8c19730133b6bc9c2b6088fd9f0ff376f9539dcf0932214b77415e542b6-image.png" alt="image.png"></p>
<p><img src="https://pic.leetcode-cn.com/1ff7a2c278f3624bae64e29bc4e3a7aeb2f1fb3835baaeb444f5d4c5df1e4c7d-image.png" alt="image.png"></p>
<p><strong>参考代码</strong>：</p>
<p>Python 代码：</p>
<pre class="line-numbers language-Python"><code class="language-Python">from typing import List


class Solution:
    def findClosestElements(self, arr: List[int], k: int, x: int) -> List[int]:
        size = len(arr)
        left = 0
        right = size - k

        while left < right:
            # mid = left + (right - left) // 2
            mid = (left + right) >> 1
            # 尝试从长度为 k + 1 的连续子区间删除一个元素
            # 从而定位左区间端点的边界值
            if x - arr[mid] > arr[mid + k] - x:
                left = mid + 1
            else:
                right = mid
        return arr[left:left + k]<span aria-hidden="true" class="line-numbers-rows"><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span></span></code></pre>
<p>Java 代码：</p>
<pre class="line-numbers language-Java"><code class="language-Java">import java.util.ArrayList;
import java.util.List;

public class Solution {

    public List<Integer> findClosestElements(int[] arr, int k, int x) {
        int size = arr.length;

        int left = 0;
        int right = size - k;

        while (left < right) {
            // int mid = left + (right - left) / 2;
            int mid = (left + right) >>> 1;
            // 尝试从长度为 k + 1 的连续子区间删除一个元素
            // 从而定位左区间端点的边界值
            if (x - arr[mid] > arr[mid + k] - x) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }

        List<Integer> res = new ArrayList<>();
        for (int i = left; i < left + k; i++) {
            res.add(arr[i]);
        }
        return res;
    }

    public static void main(String[] args) {
        int[] arr = {0, 0, 1, 2, 3, 3, 4, 7, 7, 8};
        int k = 3;
        int x = 5;
        Solution2 solution = new Solution2();
        List<Integer> res = solution.findClosestElements(arr, k, x);
        System.out.println(res);
    }
}<span aria-hidden="true" class="line-numbers-rows"><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span><span></span></span></code></pre>
<p><strong>复杂度分析</strong>：</p>
<ul>
<li>时间复杂度：$O(\log N)$，这里 $N$ 是数组的长度，使用二分法的时间复杂度是对数级别的。</li>
<li>空间复杂度：$O(1)$，只使用了常数个额外的辅助空间。</li>
</ul>
<p>（本节完）</p>

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                            「力扣」第 704 题：二分查找（简单）
链接


给定一个 n 个元素有序的（升序）整型数组 nums 和一个目标值 target ，写一个函数搜索 nums 中的 target，如果目标值存在返回下标，否则返回 -1。
示例 1：
输入
                        
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                                    专题 2：二分查找
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                        <span class="chip bg-color">减治算法</span>
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                            「力扣」第 611 题：有效三角形的个数题解地址：二分查找 （Python 代码、Java 代码）。
说明：文本首发在力扣的题解版块，更新也会在第 1 时间在上面的网站中更新，这篇文章只是上面的文章的一个快照，您可以点击上面的链接看到其他网
                        
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                                    专题 2：二分查找
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                        <span class="chip bg-color">减治算法</span>
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